Question

If each balloon is filled with carbon dioxide gas at 20ºC and 1 atmosphere, calculate the mass and the number of moles of carbon dioxide in each balloon at maximum inflation. Use the ideal gas law in your calculation

Answer 1

ideal gas law:\[PV = nRT\]where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles, \(R\) is gas law constant in appropriate units, and \(T\) is temperature in degree Kelvin. You haven't shared the entire problem statement with us, so we can't really do much more than that.

Answer 2

whats missing

Answer 3

That table of data was missing!

So rather than do the problem for you, I will make up a 4th balloon and do the work for it:

D 4 tablets circumference 27 cm radius 4.30 volume 333 cm^3

We know the pressure is 1 atmosphere.
From the table, we have volume of 333 cm^3
Temperature is given as 20 C = 20 + 273 = 293 K
ideal gas constant for atmospheres is 0.082057338 L atm / (mol K)

\[PV = nRT\]

we want to find \(n\) for the balloon, and from the number of moles we can find the mass.

\[PV = nRT\]\[n = \frac{PV}{RT} = \frac{1\text{ atm} * 333\text{ cm}^3*(1 \text{ liter}/1000\text{ cm}^3)}{(0.082057338 \text{ L atm/(mol K)})*(293\text{ K})}\]\[n = 0.0139\text{ mol}\]

molecular mass of CO2 is \[12.0107 \text{ g/mol} + 2*15.9994\text{ g/mol} = 44.0095\text{ g/mol}\]
mass of that many grams of CO2 is
\[0.0139\text{ mol}*44.0095\text{ g/mol} = 0.610\text{ g CO2}\]

Answer 4

ok i sorta get it so when calculating the mass what numbers do you multiply i still dont get that part

Answer 5

you multiply the number of moles (\(n\)) by the mass of one mole. I found the mass of one mole of CO2 for you — 44.0095 grams.

Answer 6

so in the equation what stays and what do i substitute with my info

Answer 7

any number that came from my table entry is obviously going to change. what all is in the table? volume. what is fixed by the problem? pressure and temperature. what is a physical constant? gas law constant? what do you have to find? pressure.

Answer 8

im confused lol i feels dumb

Answer 9

do you see the formula I used to find the pressure?

Answer 10

plug in the three different volumes from your table and get 3 different numbers of moles of CO2 for the three balloons. then convert each of those numbers of moles to grams of CO2 by multiplying by the molecular mass of 1 mole of CO2, which is 44.0095 grams, as previously stated. That is it. You just need to do the identical problem 3 times.

Answer 11

no

Answer 12

it lagged and said no to late lol

Answer 13

sorry, the number of moles, not the pressure.

Answer 14

what is the volume for balloon A?

Answer 15

i dont know how the table works
n=PVRT=1 atm∗333 cm3∗(1 liter/1000 cm3)
----------------------------------------
(0.082057338 L atm/(mol K))∗(293 K)

Answer 16

like i dont remember what cancels each other out or what im supposed to multiply

Answer 17

:P im guessing you are in connexus im doing this exact same project right now ^^

Answer 18

@whpalmer4 this helped a lot for me thanx so much ^^

Answer 19

i got it

Answer 20

So the table gave us the volume for each balloon, in the final column.

My made-up balloon had a volume of \(333 \text{ cm}^3\) and the volume is the only parameter that changes between the balloons.

\[PV = nRT\]we want to find the value of \(n\) for each of the balloons.
\[\frac{PV}{RT} = \frac{nRT}{RT}\]\[\frac{PV}{RT} = \frac{n\cancel{RT}}{\cancel{RT}}\]\[n = \frac{PV}{RT}\]\[1\text{ L} = 1000 \text{ cm}^3\]

\[P = 1 \text{ atm}\]\[T=293^\circ\text{K}\]\[R=0.082057338\frac{\text{L atm}}{\text{mol }^\circ\text{K}}\]

\[n= \frac{1\text{ atm} * 333 \text{ cm}^3*\frac{1 \text{ L}}{1000\text{ cm}^3}}{0.082057338\frac{\text{L atm}}{\text{mol }^\circ\text{K}} * 293 ^\circ\text{K} }\]
\[n= \frac{1\text{ atm} * 333 \cancel{\text{ cm}^3}*\frac{1 \text{ L}}{1000\cancel{\text{ cm}^3}}}{0.082057338\frac{\text{L atm}}{\text{mol }^\circ\text{K}} * 293 ^\circ\text{K} }\]\[n= \frac{1\cancel{\text{ atm}} * 333 \cancel{\text{ cm}^3}*\frac{1 \text{ L}}{1000\cancel{\text{ cm}^3}}}{0.082057338\frac{\text{L }\cancel{\text{atm}}}{\text{mol }^\circ\text{K}} * 293 ^\circ\text{K} }\]\[n= \frac{1\cancel{\text{ atm}} * 333 \cancel{\text{ cm}^3}*\frac{1 \cancel{\text{ L}}}{1000\cancel{\text{ cm}^3}}}{0.082057338\frac{\cancel{\text{L }}\cancel{\text{atm}}}{\text{mol }^\circ\text{K}} * 293 ^\circ\text{K} }\]\[n= \frac{1\cancel{\text{ atm}} * 333 \cancel{\text{ cm}^3}*\frac{1 \cancel{\text{ L}}}{1000\cancel{\text{ cm}^3}}}{0.082057338\frac{\cancel{\text{L }}\cancel{\text{atm}}}{\text{mol }\cancel{^\circ\text{K}}} * 293 \cancel{^\circ\text{K} }}\]\[n=\frac{1*333}{0.082057338*1000*293}\text{ mol} = 0.0139 \text{ mol} \]

For a problem with a different volume, just put the new volume in place of the \(333\) in that last calculation.